Chapter 5, Problem 81GQ

The standard molar enthalpy of formation of diborane, B₂H₆(g), cannot be determined directly because the compound cannot be prepared by the reaction of boron and hydrogen. It can be calculated from other enthalpy changes, however. The following enthalpy changes can be measured.

4 B(s) + 3 O₂(g) → 2 B₂O₃(s)

Δ_rH° = –2543.8 kJ/mol-rxn

H₂(g) +½ O₂(g) → H₂O(g)

Δ_rH° = –241.8 kl/mol-rxn

B₂H₆(g) + 3 O₂(g) → B₂O₃(s) + 3 H₂O(g)

Δ_rH° = –2032.9 kJ/mol-rxn

(a) Show how these equations can be added together to give the equation for the formation of B₂H₆(g) from B(s) and H₂(g) in their standard states. Assign enthalpy changes to each reaction.

(b) Calculate Δ_fH° for B₂H₆(g).

(c) Draw an energy level diagram that shows how the various enthalpies in this problem are related.

(d) Is the formation of B₂H₆(g) from its elements exo- or endothermic?

Interpretation Introduction

Interpretation:

The equation for the formation of ${\text{B}}_{\text{2}}{\text{H}}_{\text{6}}$ from $\text{B}\left(\text{s}\right)\text{and}{\text{H}}_{\text{2}}\left(\text{g}\right)$ has to be determined

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

First equation is:

$\text{4B}\text{+}\text{3}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{2B}}_{\text{2}}{\text{O}}_{\text{3}}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}\text{=-2543}\text{.8}\text{kJ/mol}$

Divide the 1st equation by 2

$\text{2B}{\text{+3/2O}}_{\text{2}}\to {\text{B}}_{\text{2}}{\text{O}}_{\text{3}}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}\text{=-1271}\text{.9kJ/mol}$

Multiply the 2^nd equation by 3

${\text{H}}_{\text{2}}\text{+}1{\text{/2O}}_{\text{2}}\stackrel{}{\to }{\text{H}}_{\text{2}}\text{O}{\text{ΔH}}^{\text{0}}\text{=}\text{-241}\text{.8kJ/mol}$

${\text{3H}}_{\text{2}}\text{+}{\text{3/2O}}_{\text{2}}\stackrel{}{\to }{\text{3H}}_{\text{2}}\text{O}{\text{ΔH}}^{\text{0}}\text{=-725}\text{.4kJ/mol}$

Reverse the 3^rd equation and the obtained equation is:

${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}{\text{3H}}_{\text{2}}\text{O}\to {\text{B}}_{\text{2}}{\text{H}}_{\text{6}}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}\text{=}\text{2032}\text{.9kJ/mol}$

Interpretation Introduction

Interpretation:

The formation energy for ${\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\left(\text{g}\right)$ has to be determined.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

First equation is:

$\text{4B}\text{+}\text{3}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{2B}}_{\text{2}}{\text{O}}_{\text{3}}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}\text{=-2543}\text{.8}\text{kJ/mol}$

Divide the 1st equation by 2

$\text{2B}{\text{+3/2O}}_{\text{2}}\to {\text{B}}_{\text{2}}{\text{O}}_{\text{3}}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}\text{=-1271}\text{.9kJ/mol}$

Second equation is:

${\text{H}}_{\text{2}}\text{+}1{\text{/2O}}_{\text{2}}\stackrel{}{\to }{\text{H}}_{\text{2}}\text{O}{\text{ΔH}}^{\text{0}}\text{=}\text{-241}\text{.8kJ/mol}$

Multiply the 2^nd equation by 3

${\text{3H}}_{\text{2}}\text{+}{\text{3/2O}}_{\text{2}}\stackrel{}{\to }{\text{3H}}_{\text{2}}\text{O}{\text{ΔH}}^{\text{0}}\text{=-725}\text{.4kJ/mol}$

And the third equation is:

${\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\text{+}{\text{3H}}_{\text{2}}\text{O}\to {\text{B}}_{\text{2}}{\text{H}}_{\text{6}}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}\text{=}\text{2032}\text{.9kJ/mol}$

Adding the 3 equations gives the required equation.

$\text{2B}\text{+}{\text{3H}}_{\text{2}}\to {\text{B}}_{\text{2}}{\text{H}}_{\text{6}}\text{+}{\text{3/2O}}_{\text{2}}{\text{ΔH}}^{\text{0}}\text{=35}\text{.6kJ}$

So, the formation constant is $\text{35}\text{.6}\text{kJ/mol}$.

Interpretation Introduction

Interpretation:

The energy level diagram for the various enthalpies in the problem has to be depicted.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

The energy level diagram is shown below.

Figure 1

Interpretation Introduction

Interpretation:

The nature of the thermodynamic reaction has to be identified.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

When heat energy is transferred the directionality of heat transfer between the system and the surrounding is described either as endothermic or exothermic

Energy of the system decrease and energy of the surrounding increases, i.e ${\text{q}}_{\text{sys}}\text{<0,}$ Energy is transferred as heat from the surroundings to the system and work is done by the system -exothermic.

Endothermic process, the heat energy is transferred from surroundings to the system. Energy of system increases and that of surroundings decreases i.e, ${\text{q}}_{\text{sys}}\text{>0}$

Explanation of Solution

The formation constant is $\text{+35}\text{.6}\text{kJ/mol}$, so the reaction is endothermic in nature.